course Mth 158 \PrxEޠiH{assignment #013
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20:50:44 **** query 1.5.34 (was 1.5.24). How did you write the interval [0, 1) using an inequality with x? Describe your illustration using the number line.
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RESPONSE --> On the number line you would put a dot that is not filled in above the 0 and draw a bold line to the 1 with a dot that is not filled in above the 1.
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20:51:27 ** My notes here show the half-closed interval [0, 1). When sketching the graph you would use a filled dot at x = 0 and an unfilled dot at x = 1, and you would fill in the line from x = 0 to x = 1. **
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RESPONSE --> Oh okay so at the zero you would have a filled dot. I understand that.
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20:52:23 **** query 1.5.40 (was 1.5.30). How did you fill in the blank for 'if x < -4 then x + 4 ____ 0'?
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RESPONSE --> x + 4 ___ 0 x < -4
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20:52:40 ** if x<-4 then x cannot be -4 and x+4 < 0. Algebraically, adding 4 to both sides of x < -4 gives us x + 4 < 0. **
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RESPONSE --> ok
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20:53:03 **** query 1.5.46 (was 1.5.36). How did you fill in the blank for 'if x > -2 then -4x ____ 8'?
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RESPONSE --> -4x ___ 8 x < 8
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20:53:05 **if x> -2 then if we multiply both sides by -4 we get -4x <8. Recall that the inequality sign has to reverse if you multiply or divide by a negative quantity. **
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RESPONSE -->
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20:53:57 **** query 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1.
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RESPONSE --> 2x + 5 >= 1 2x >= -4 x <= -2
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20:54:41 ** Starting with 2x+5>= 1 we add -5 to both sides to get 2x>= -4, the divide both sides by 2 to get the solution x >= -2. **
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RESPONSE --> okay, why would you not change the sign on this one, because you had to divide both sides by 2
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20:56:39 **** query 1.5.64 (was 1.5.54). Explain how you solved the inquality 8 - 4(2-x) <= 2x.
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RESPONSE --> 8 - 4(2 - x) <= 2x 8 - 8 - (-4x) <= 2x 4x <= 2x 4x - 2x <= 0 2x <= 0 x <= 0
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20:56:45 ** 8- 4(2-x)<= 2x. Using the distributive law: 8-8+4x<= 2x. Simplifying: 4x<=2x. Subtracting 2x from both sides: 2x<= 0 x<=0 **
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RESPONSE --> ok
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20:57:49 **** query 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1.
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RESPONSE --> 0 < 1 - 1/3x < 1 -1 < -1/3x < 0 3 < x < 0 or 0 < x < 3
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20:57:57 ** Starting with 0<1- 1/3x<1 we can separate this into two inequalities, both of which must hold: 0< 1- 1/3x and 1- 1/3x < 1. Subtracting 1 from both sides we get -1< -1/3x and -1/3x < 0. We solve these inequalitites separately: -1 < -1/3 x can be multiplied by -3 to get 3 > x (multiplication by the negative reverses the direction of the inequality) -1/3 x < 0 can be multiplied by -3 to get x > 0. So our inequality can be written 3 > x > 0. This is not incorrect but we usually write such inequalities from left to right, as they would be seen on a number line. The same inequality is expressed as 0 < x < 3. **
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RESPONSE --> ok
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20:59:45 **** query 1.5.94 (was 1.5.84). Explain how you found a and b for the conditions 'if -3 < x < 3 then a < 1 - 2x < b.
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RESPONSE --> 6 < 1 - 2x < -6 7 < 1 - 2x < -5 a = 7 b = -5
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21:00:19 ** Adding 1 to each expression gives us 1 + 6 > 1 - 2x > 1 - 6, which we simplify to get 7 > 1 - 2x > -5. Writing in the more traditional 'left-toright' order: -5 < 1 - 2x < 7. **
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RESPONSE --> ok, I just didn't write it in the traditional order
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21:01:51 **** query 1.5.106 (was 1.5.96). Explain how you set up and solved an inequality for the problem. Include your inequality and the reasoning you used to develop the inequality. Problem (note that this statement is for instructor reference; the full statement was in your text) commision $25 + 40% of excess over owner cost; range is $70 to $300 over owner cost. What is range of commission on a sale?
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RESPONSE --> 70 < 0.40x < 300 0.40 * 70 < 0.40x < 0.40 * 300 25 + 28 < 25 + 0.40x < 25 + 120 53 < 25 + 0.40x < 145
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21:01:58 ** If x = owner cost then 70 < x < 300. .40 * owner cost is then in the range .40 * 70 < .40 x < .40 * 300 and $25 + 40% of owner cost is in the range 25 + .40 * 70 < 25 + .40 x < 25 + .40 * 300 or 25 + 28 < 25 + .40 x < 25 + 120 or 53 < 25 + .40 x < 145. **
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RESPONSE --> ok
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21:02:43 **** query 1.5.112. Why does the inequality x^2 + 1 < -5 have no solution?
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RESPONSE --> x^2 + 1 < -5 x^2 < -4 x < sqrt -4 (cannot take the square root of a negative number)
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21:02:51 STUDENT SOLUTION: x^2 +1 < -5 x^2 < -4 x < sqrt -4 can't take the sqrt of a negative number INSTRUCTOR COMMENT: Good. Alternative: As soon as you got to the step x^2 < -4 you could have stated that there is no such x, since a square can't be negative. **
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RESPONSE --> ok
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ﹲ며{~ assignment #014 U|\ȔFQ College Algebra 10-01-2006
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21:04:59 **** query 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 0.
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RESPONSE --> | 1 - 2z | + 6 = 9 1 - 2z = 3 -2z = 2 z = -1 or 1 - 2z = -3 -2z = -4 z = 2
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21:05:50 ** Starting with | 1-2z| +6 = 9 we add -6 to both sides to get | 1 - 2z| = 3. We then use the fact that | a | = b means that a = b or a = -b: 1-2z=3 or 1-2z= -3 Solving both of these equations: -2z = 2 or -2z = -4 we get z= -1 or z = 2 We express our solution set as {-2/3,2} **
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RESPONSE --> How do you get the set (-2/3 , 2)?
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21:08:07 **** query 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2
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RESPONSE --> x^2 + 3x - 2 = 2 x^2 + 3x - 4 = 0 (x +4)(x-1) = 0 x = -4, x = 1 or x^2 + 3x - 2 = -2 x^2 + 3x = 0 x(x + 3) = 0 x = 0, x = -3
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21:08:14 ** My note here might be incorrect. If the equation is | x^2 +3x -2 | = 2 then we have x^2 + 3x - 2 = 2 or x^2 + 3x - 2 = -2. In the first case we get x^2 + 3x - 4 = 0, which factors into (x-1)(x+4) = 0 with solutions x = 1 and x = -4. In the second case we have x^2 + 3x = 0, which factors into x(x+3) = 0, with solutions x = 0 and x = -3. **
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RESPONSE --> ok
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21:09:21 **** query 1.6.36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5.
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RESPONSE --> | x + 4| + 3 < 5 | x + 4| < 2 -2 < x + 4 < 2 -6 < x < -2
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21:09:26 STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2
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RESPONSE --> ok
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21:10:48 **** query 1.6.48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1.
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RESPONSE --> | -x - 2 | >= 1 x + 2 >= -1 x >= -3 or x + 2 >= 1 x >= -1
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21:11:48 **Correct solution: | -x -2 | >= 1 Since | a | > b means a > b or a < -b (note the word 'or') we have -x-2 >= 1 or -x -2 <= -1. These inequalities are easily solved to get -x >= 3 or -x <= 1 or x <= -3 or x >= -1. So our solution is {-infinity, -3} U {-1, infinity}. **
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RESPONSE --> ok, I understand where you got the infinity I just didn't realize that I needed to use it on this equation.
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